The lifespans of tigers in a particular zoo are normally distributed. The average tiger lives $19.1$ years; the standard deviation is $3$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a tiger living less than $25.1$ years.
Answer: $19.1$ $16.1$ $22.1$ $13.1$ $25.1$ $10.1$ $28.1$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $19.1$ years. We know the standard deviation is $3$ years, so one standard deviation below the mean is $16.1$ years and one standard deviation above the mean is $22.1$ years. Two standard deviations below the mean is $13.1$ years and two standard deviations above the mean is $25.1$ years. Three standard deviations below the mean is $10.1$ years and three standard deviations above the mean is $28.1$ years. We are interested in the probability of a tiger living less than $25.1$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the tigers will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the tigers will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $13.1$ years and the other half $({2.5\%})$ will live longer than $25.1$ years. The probability of a particular tiger living less than $25.1$ years is ${95\%} + {2.5\%}$, or $97.5\%$.